Lamé parameters

Remarks

Example: with $\lambda = 1$ and $\mu = 1$ and the pure shear strain $\varepsilon = \gamma (e_1 \otimes e_2 + e_2 \otimes e_1)/2$, the trace is $0$ and $\sigma_{12} = 2\mu \cdot \gamma/2 = \gamma$; so $\mu$ is the ratio of shear stress to engineering shear strain $\gamma$.

Non-example: a cubic crystal has an elasticity tensor with three independent constants, and in general it is not of the form above, so it has no Lamé parameters.

Positivity of the elasticity tensor is equivalent to $\mu > 0$ and $3\lambda + 2\mu > 0$. In terms of Young's modulus $E$ and Poisson's ratio $\nu$, $\lambda = \frac{E \nu}{(1 + \nu)(1 - 2\nu)}$ and $\mu = \frac{E}{2(1 + \nu)}$.